Problem 24 - Algebra and Trigonometry

Question

\(19-28\) . Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. $$ a=75, \quad b=100, \quad \angle A=30^{\circ} $$

Step-by-Step Solution

Answer

Two possible triangles: \((30^{\circ}, 41.81^{\circ}, 108.19^{\circ})\) and \((30^{\circ}, 138.19^{\circ}, 11.81^{\circ})\).

Step 1: Understand the Problem

We're given side lengths \(a = 75\) and \(b = 100\) and angle \(∠A = 30^{\circ}\). We need to determine which triangles can be formed using the Law of Sines. The Law of Sines is \(\frac{a}{\sin A} = \frac{b}{\sin B}\). Our goal is finding \(\angle B\) and validating possible triangle formations.

Step 2: Apply the Law of Sines

According to the Law of Sines, we can set up the equation \(\frac{75}{\sin 30^{\circ}} = \frac{100}{\sin B}\). Since \(\sin 30^{\circ} = \frac{1}{2}\), substitute to get \(\frac{75}{\frac{1}{2}} = \frac{100}{\sin B}\). Simplifying gives \(150 = \frac{100}{\sin B}\). Solving for \(\sin B\), we have \(\sin B = \frac{100}{150}\).

Step 3: Calculate Possible Values of Angle B

From \(\sin B = \frac{2}{3}\), we find \(\angle B\) using inverse sine: \(B = \sin^{-1}(\frac{2}{3}) \approx 41.81^{\circ}\). Since the sine function gives two possible angles (\(B\) and \(180^{\circ} - B\)), check for another possibility: \(180^{\circ} - 41.81^{\circ} = 138.19^{\circ}\). Both \(41.81^{\circ}\) and \(138.19^{\circ}\) should be tested if they fulfill triangle conditions.

Step 4: Check Triangle Validity

For \(B = 41.81^{\circ}\), calculate \(∠C = 180^{\circ} - 30^{\circ} - 41.81^{\circ} = 108.19^{\circ}\), which is valid since all angles add to \(180^{\circ}\). Check for \(B = 138.19^{\circ}\), calculate \(∠C = 180^{\circ} - 30^{\circ} - 138.19^{\circ} = 11.81^{\circ}\), valid as well. Therefore, two triangles are possible with angles \(30^{\circ}, 41.81^{\circ}, 108.19^{\circ}\) and \(30^{\circ}, 138.19^{\circ}, 11.81^{\circ}\).

Step 5: Conclusion

Based on calculations, two triangles are possible. Triangle 1: \(∠A = 30^{\circ}\), \(∠B = 41.81^{\circ}\), \(∠C = 108.19^{\circ}\); Triangle 2: \(∠A = 30^{\circ}\), \(∠B = 138.19^{\circ}\), \(∠C = 11.81^{\circ}\). Both configurations satisfy triangle angle sum and side conditions.

Key Concepts

Solving Triangles

When it comes to solving triangles, especially non-right triangles, understanding the relationships between sides and angles is crucial. One effective tool for this is the Law of Sines. This law connects the lengths of sides of a triangle to the sines of its opposite angles. The formula is simple yet powerful: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]In practical terms, this means that if we know some parts of a triangle, like two sides and a non-included angle (known as the SSA condition), we can usually solve for the missing parts. A step-by-step approach involves:

• Identifying known values (sides and angle).
• Applying the Law of Sines to set up an equation.
• Calculating the unknowns, such as an angle or a third side.
• Verifying the validity of the triangle's properties (such as checking if the sum of angles equals 180°).

Solving triangles can sometimes lead to multiple solutions, as demonstrated in cases where the sine function provides more than one possible angle. Recognizing these possibilities is key to fully understanding the problem.

Angle Calculation

Calculating angles in triangles involves not just arithmetic, but also a bit of geometry know-how. When using the Law of Sines, starting with at least one angle is essential as it allows the calculation of other angles. Given the problem where side \( a = 75 \), side \( b = 100 \), and angle \( \angle A = 30^{\circ} \), the challenge was to find \( \angle B \) using \[ \frac{a}{\sin A} = \frac{b}{\sin B} \]Substituting in the known values, you find \[ \frac{75}{\sin 30^{\circ}} = \frac{100}{\sin B} \]so \( \sin B \) is calculated and gives two potential values for \( \angle B \). The first, from directly computing the inverse sine, often gives an acute angle. But because \[ \sin(\theta) = \sin(180^{\circ} - \theta) \]we can also consider the obtuse angle complementary scenario. This is why trigonometric problems can sometimes yield more than one triangular configuration.During angle calculation, a final check ensures the internal angles are valid, adding up to exactly 180 degrees.

Trigonometric Identities

Utilizing trigonometric identities is an integral part of angle calculation and solving triangles. These identities simplify and guide the solution process involving angles. In the Law of Sines, the primary trigonometric function in use is the sine function, known for its periodic properties. Key trigonometric identities that might help further understand angles include:

• \( \sin(180^{\circ} - \theta) = \sin(\theta) \)
• \( \cos(\theta) = \cos(360^{\circ} - \theta) \)
• Sine and Cosine Pythagorean Identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

In the problem context, recognizing that \( B \) could have two values, \( 41.81^{\circ} \) and \( 138.19^{\circ} \), is rooted in the property \( \sin \) retains the same value within one period, or 180 degrees.Through trigonometric identities, one can evaluate and combine angles meticulously to ensure all possible triangle solutions are considered. They also confirm that the angles make logical sense within the geometric construct of the problem at hand.